14 INTEGRATION WITH RESPECT TO A POSITIVE MEASURE 187 take the values 0 and ±00 simultaneously, it follows from (13.9.8.1) that (b) implies (c). Conversely, it is immediately seen that the function gf which is equal to g~l on S and vanishes on X — S is //-measurable : for there exists a partition of S (resp. X — S) into compact sets Ln (resp. Mn) and a //-negligible set P (resp. Q) such that the restriction of g to each Ln is continuous, and it follows that the restrictions of g' to Ln and Mn are continuous. Furthermore, we have (fg)gf =fty$ , hence by the same argument it follows that (c) implies (b). Hence it is enough to prove that (a) and (b) are equivalent. Suppose first that/(ps is //-measurable. The set (JS is v-negligible, by virtue of (1 3.14.1 .5) ; also the hypothesis implies the existence of a partition of S into a sequence of compact sets LM and a /(-negligible set N, such that f\ Ln is continuous for each n. Also N is v-negligible, by (13.14.1.3); hence / is v-measurable. Conversely, suppose that/is v-measurable. Then there exists a partition of X consisting of a sequence of compact sets Hn and a v-negligible set N, such that/I Hn is continuous for all n. Also there exists a partition of S consisting of a //-negligible set L and a sequence of compact sets Kn such that g \ Kn is con- tinuous (and >0) for all n. For all «, we have inf g(x) = an > 0. Finally, xeKn there exists also a partition of X — S into a /(-negligible set L' and a sequence of compact sets K^ . It is clear that the restriction of fys to each of the sets Hn n Km and K'm is continuous, and it is therefore enough to prove that N n S is //-negligible. Now if //*(N n Krt) > 0, then we should have ^ an //*(N n Kn) > 0, which is absurd because JNnK 0 = V(N n S) ~JN nS by virtue of (13.14.1). Hence each set N n Kn is //-negligible, therefore so is N n S, and hence finally /<ps is //-measurable. (13.14.3) Let g be a locally //-//?tegrable function which is ^0 on X, and let v = g - //. Then a mapping f: X -> K is v-integrable if and only if (with the con- ventions of (13.11))^ is \i-integr able \ and in that case we have (13.14.3.1) \fdv= \fgdjA. For/to be v-integrable it is necessary and sufficient that/4" and/"" should be (13.7.4); and since we have(jfc)+ ~f+g, (fg)~ =/"#, with the conventions of 13.11 about products, it is enough to prove (13.14.3) when/^ 0. The first assertion is now a consequence of (13.14.1), (13.14.2), and (13.9.13); and the relation (13.14.3.1) is just (13.14.1.1). w))~ |2,